Let $\Omega \subseteq \Bbb R^n$ be an open set and $K \subseteq \Omega$ compact. Prove that there exists an $r \gt 0$ such that the set $$\left \{y \in \Bbb R^n\ :\ \|y-x\| \leq r\ \text {for some}\ x \in K \right \}$$ is a compact subset of $\Omega.$
My attempt $:$
Let $x \in K.$ Then there exists $r_x \gt 0$ such that $B(x,r_x) \subseteq \Omega.$ Consider the collection $$\mathcal C : =\left \{B \left (x,\dfrac {r_x} {2} \right )\ :\ x \in K \right \}.$$ Then $\mathcal C$ is an open cover of $K$ and hence it has finite subcover say $$\mathcal C' = \left \{B \left (x_i,\dfrac {r_i} {2} \right )\ :\ 1 \leq i \leq n \right \}$$ where $r_i = r_{x_i},$ for $i =1,2, \cdots ,n.$ Let $r = \min \left \{\dfrac {r_i} {2}\ :\ 1 \leq i \leq n \right \}.$ Define a set $$S : = \left \{y \in \Bbb R^n\ :\ \|y-x\| \leq r\ \text {for some}\ x \in K \right \}$$ We will first show that $S \subseteq \Omega.$ For that let us take $y \in S.$ Then there exists $x \in K$ such that $\|y - x \| \leq r.$ Since $x \in K$ there exists $1 \leq i \leq n$ such that $x \in B \left (x_i, \dfrac {r_i} {2} \right ).$ But then by triangle inequality we have $$\|y - x_i\| \leq \|y - x\| + \|x - x_i\| \lt r + \dfrac {r_i} {2} \leq \dfrac {r_i} {2} + \dfrac {r_i} {2} = r_i.$$ This shows that $y \in B(x_i,r_i) \subseteq \Omega.$ This proves that $S \subseteq \Omega.$
Now we claim that $S$ is closed. To prove that let us take $\varepsilon \gt 0$ arbitrarily. Let $z$ be an accumulation point of $S.$ Then there exists a sequence $\{y^m\}_{m \geq 1}$ in $S$ converging to $z.$ Since $y^m$'s are in $S$ there exists a sequence $\{x^m\}_{m \geq 1}$ in $K$ such that $\|y^m - x^m\| \leq r.$ Since $K$ is compact there exists a convergent subsequence of $\{x^m\}_{m \geq 1}$ say $\{x^{m_k}\}_{k \geq 1}.$ Let $x_0 \in K$ be it's limit. Then we can find $N \in \Bbb N$ such that for all $k \geq N$ we have \begin{align*} \|y^{m_k} - z \| & \lt \dfrac {\varepsilon} {2}; \\ \|x^{m_k} - x_0\| & \lt \dfrac {\varepsilon} {2}. \end{align*} But then by triangle inequality we have $$\|z - x_0\| \leq \|z - y^{m_N}\| + \|y^{m_N} - x^{m_N}\| + \|x^{m_N} - x_0\| \lt \dfrac {\varepsilon} {2} + r + \dfrac {\varepsilon} {2} = r + \varepsilon.$$ Since $\varepsilon > 0$ was arbitrarily taken it follows that $\|z - x_0\| \leq r,$ proving that $z \in S.$ This shows that $S$ is closed.
Now we will show that $S$ is bounded. For that let us take any two elements $y',y'' \in S.$ Then there exist $x',x'' \in K$ such that $\|y' - x'\| \leq r$ and $\|y'' - x''\| \leq r.$ Since $x',x'' \in K,$ there exist $1 \leq i,j \leq n$ such that $\|x' - x_i\| \lt \dfrac {r_i} {2}$ and $\|x'' - x_j\| \lt \dfrac {r_j} {2}.$ Let $M = \max \left \{\|x_i - x_j\|\ :\ 1 \leq i \lt j \leq n \right \}$ and $s = \max \left \{r_i\ :\ 1 \leq i \leq n \right \}.$ Then by triangle inequality we have $$\|x' - x''\| \leq \|x' - x_i\| + \|x_i - x_j\| + \|x_j - x''\| \lt \dfrac {r_i} {2} + M + \dfrac {r_j} {2} \leq \dfrac {s} {2} + M + \dfrac {s} {2} = M + s.$$ But then again by triangle inequality we have $$\|y' - y''\| \leq \|y' - x'\| + \|x' - x''\| + \|x'' - y''\| \lt r + M + s + r = M + s + 2r.$$ Let $R = M + s + 2r.$ Then clearly $S \subseteq B(y,R),$ for any $y \in S.$ This shows that $S$ is bounded.
Hence $S \subseteq \Omega \subseteq \Bbb R^n$ is a closed and bounded subset and hence it is a compact subset of $\Omega,$ by Heine-Borel theorem on $\Bbb R^n.$ This completes the proof.
QED
Can anybody please check my proof whether it holds good or not? Thanks for reading.
